Misc 15 - Chapter 6 Class 12 Application of Derivatives

Transcript

Misc 15 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4 3 . Given, Radius of sphere = r Let R be the radius of the cone and H be its height. Let BOC = Now, AC = AO + OC H = r + r cos H = r (1 + cos ) And, R = r sin We need to maximize volume of cone. Volume of the cone is V = 1 3 2 V = 1 3 2 sin 2 r (1 + cos ) V = 1 3 3 sin 2 (1 + cos ) Finding = 1 3 3 2 sin cos (1+ cos )+ sin 2 ( sin ) = 1 3 3 2 sin cos (1+ cos ) sin 3 = 1 3 3 2 sin cos (1+ cos ) sin 3 = 1 3 3 sin 2 cos + 2 cos 2 sin 2 = 1 3 3 sin 2 cos + 2 cos 2 (1 cos 2 ) = 1 3 3 sin 3 cos 2 + 2 1 = 1 3 3 sin 3 cos 2 + 3 cos 1 = 1 3 3 sin 3 cos +1 1 cos +1 = 1 3 3 sin 3 1 +1 Putting = 0 1 3 3 sin +1 3 1 = 0 sin (cos + 1) (3 cos 1) = 0 So, = 1 3 Thus, H = r (1 + cos ) H = r 1 + 1 3 H = 4 3 This value is either maxima or minima. So, finding 2 2 = 1 3 3 sin 3 cos2 + 2 cos 1 = 1 3 3 cos 3 cos 2 +2 cos 1 + sin 6 cos 2 sin = 1 3 3 3 cos 3 +2 cos 2 cos + sin 6 cos sin 2 sin = 1 3 3 3 cos 2 +2 cos 2 cos 6 sin 2 cos 2 sin 2 Now, cos = 1 3 And sin2 = 1 cos2 = 1 1 3 2 = 1 1 9 = 8 9 Putting values in 2 2 2 2 = 1 3 3 3 1 3 3 +2 1 3 2 1 3 6 1 3 8 9 2 8 9 = 1 3 3 3 1 27 +2 1 9 1 3 6 8 27 2 8 9 = 1 3 3 1 9 + 2 9 1 3 16 9 16 9 = 1 3 3 32 9 = 32 3 27 Thus, 2 2 < 0 H = Hence proved